Physics

Ray Optics and Mirrors

141 Questions

Ray optics covers the principles of light reflection and refraction through mirrors and lenses. The questions focus on focal length, magnification, and image formation. This physics topic is highly relevant for exams requiring science aptitude.

Concave mirror imagesConvex lens formulasFocal length calculationsMagnification ratiosImage distance

Ray Optics and Mirrors Questions

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

An object is placed on the principal axis of a concave mirror at a distance of 60 cm.  If the focal length of the concave mirror is 40 cm then determine the magnification of the obtained image.

  1. 4

  2. -2

  3. -4

  4. +2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given: $u=-60 cm$     $f=-40 cm$
To find : $m$
Solution: From mirror formula
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{1}{v}-\dfrac{1}{60}=\dfrac{-1}{40}$
$\dfrac{1}{v}=\dfrac{1}{60}-\dfrac{1}{40}$
$v=-120 cm$
Hence magnification is given by
$m=\dfrac{-v}{u}=-\dfrac{(-120)}{(-60)}$
$m=-2$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

Calculate the magnification of an object if it is kept at a distance of $3 cm$ from a concave mirror of focal length $4 cm$:

  1. $3$
  2. $6$
  3. $9$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

$\dfrac{1}{-3}+\dfrac{1}{v}=\dfrac{1}{-4}$
$\dfrac{1}{v}=\dfrac{1}{12}$
v= 12 cm
Magnification
$=\dfrac{12}{3}=4$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

In an experiment to determine the focal length ($f$) of a concave mirror by the $u-v$ method, a student places the object pin A on the principal axis at a distance $x$ from the pole $P$. The student looks at the pin and its inverted image from a distance keeping the eye in line with $PA$. When the student shifts the eye towards left, the image appears to the right of the object pin. Then:

  1. $x< f$
  2. $f< x< 2f$
  3. $x= 2f$
  4. $x> 2f$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the object and the image move in opposite directions, the position of the object should be in between $f$ and $2f$.
So, $f < x < 2f$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A concave mirror forms the real image of an object which is magnified 4 times. The objects is moved 3 cm away, the magnification of the image is 3 times. What is the focal length of the mirror?

  1. 3 cm

  2. 4 cm

  3. 12 cm

  4. 36 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
For mirror $u=\frac {f(m-1)}{m}$
In first case, $u=\frac {f(-4-1)}{-4}$
In the second case, $u+3=\frac {f(-3-1)}{-3}$
On solving, we get $f=36 cm$
Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

The distance between an object and its doubly magnified image by a concave mirror is: [ Assume $f$ = focal length]

  1. $ 3 f/2 $
  2. $2 f/3 $
  3. $3f$

  4. Depends on whether the image is real or virtual.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The magnification is given as,

$m = \dfrac{{ - v}}{u}$

$2 = \dfrac{{ - v}}{u}$

$v =  - 2u$

Ignoring the sign and using mirror formula, we get

$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{{1 + 2}}{{2u}} = \dfrac{1}{f}$

$u = \dfrac{{3f}}{2}$

Here, difference between object distance and image distance is also$u$.

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A short linear object of length $b$ lies along the axis of a concave mirror of focal length $f$ at a distance u from the pole of the mirror. The size of the image is approximately equal to :

  1. $b\left (\dfrac {u-f}{f}\right )^{\dfrac {1}{2}}$
  2. $b\left (\dfrac {b}{u-f}\right )^{\dfrac {1}{2}}$
  3. $b\left (\dfrac {u-f}{f}\right )$
  4. $b\left (\dfrac {f}{u-f}\right )^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From mirror formula,


$\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \longrightarrow (1)$

Differentiating, we get  


$\Rightarrow -{ \upsilon  }^{ -2 }dv-{ u }^{ -2 }du=0$

or $\left| d\upsilon  \right| =\left| \cfrac { { \upsilon  }^{ 2 } }{ { u }^{ 2 } }  \right| du \ \longrightarrow (2)$         

Here $\left| dv \right| =$size of image,

$\left| du \right| =$size of object $\left( =b \right) $

From the equation $1$, we write

$\cfrac { u }{ v } +1 =\cfrac { u }{ f } $

Squaring both sides, we get

$\cfrac { { \upsilon  }^{ 2 } }{ { v }^{ 2 } } ={ \left( \cfrac { f }{ u-f }  \right)  }^{ 2 }$

Substituting in equation $2$ we get

Size of the image  $dv=b{ \left( \cfrac { f }{ u-f }  \right)  }^{ 2 }$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The focal length of a mirror is given by $\dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$. If equal errors ($\alpha$) are made in measuring $u$ and $v$, then the relative error in $f$ is

  1. $\dfrac {2}{\alpha}$
  2. $\alpha \left (\dfrac {1}{u}+\dfrac {1}{v}\right )$
  3. $\alpha \left (\dfrac {1}{u}-\dfrac {1}{v}\right )$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $\displaystyle \dfrac {1}{v}-\dfrac {1}{u}=\dfrac {2}{f}$
$\Rightarrow \displaystyle -\dfrac {\Delta v}{v^2}+\dfrac {\Delta u}{u^2}=-\dfrac {2\Delta f}{f^2}$
Since, given equal errors in measuring u and v i.e.$\Delta u=\Delta v=\alpha$
$\Rightarrow {\alpha}\left(\dfrac{1}{u}-\dfrac{1}{v}\right)\left(\dfrac{1}{u}+\dfrac{1}{v}\right)=-\dfrac {2\Delta f}{f^2}$
But, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{2}{f}$
$\Rightarrow\displaystyle \frac{ \Delta f}{f}={\alpha}\left(\dfrac{1}{u}+\dfrac{1}{v}\right)$
Hence, correct option is B

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

A diverging lens of focal length $-10cm$ is moving towards right with a velocity $5m/s$. An object, placed on principal axis is moving towards left with a velocity $3m/s$. The velocity of image at the instant when the lateral magnification produced is $1/2$ is: (All velocities are with respect to ground)

  1. $3m/s$ towards rigtht
  2. $3m/s$ towards left
  3. $7m/s$ towards rigtht
  4. $7m/s$ towards left
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the lens formula 1/v - 1/u = 1/f and magnification m = v/u = 1/2. So v = u/2. 2/u - 1/u = -1/10, so 1/u = -1/10, u = -10, v = -5. Velocity of image v_i = m^2 * v_o = (1/2)^2 * 3 = 0.75. Considering relative velocities, the result is 3m/s towards the right.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

Which values for $K, L, M$ and $N$ will make the following paragraph true?
When an object of size $7\ cm$ is placed at the distance of $K$ in front of a $L$ of focal length $M,$ the image will be produced at the distance of $N$ in front of the mirror.

  1. $\mathrm { K } - 27\ \mathrm { cm } ; \mathrm { L-concave \ mirror}; \mathrm { M } - 18\ \mathrm { cm } ; \mathrm { N } - 36\ \mathrm { cm }$
  2. $\mathrm { K } - 18\ \mathrm { cm } ; \mathrm { L-concave\ mirror}; \mathrm { M } - 18\ \mathrm { cm } ; N - 54\ \mathrm { cm }$
  3. $\mathrm { K } - 27\ \mathrm { cm } ; L-concave\ mirror; \mathrm { M } - 18\ \mathrm { cm } ; N - 54\ \mathrm { cm }$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the mirror formula 1/v + 1/u = 1/f. For K=27, M=18 (f=-18), 1/v - 1/27 = -1/18. 1/v = 1/27 - 1/18 = (2-3)/54 = -1/54. v = -54.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

A point object is placed at a distance of $15 cm$ from a convex lens. The image is formed on the other side at a distance of $30cm$ from the lens. When a concave lens is placed in contact with the convex lens, the image shifts away further by $30 cm$. Calculate the focal lengths of the concave and convex lenses.

  1. $10 cm, 60 cm$
  2. $ 20 cm, 30 cm$
  3. $60 cm, 10 cm$
  4. $30 cm, 20 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the convex lens, 1/v - 1/u = 1/f. 1/30 - 1/-15 = 1/f1 => 1/30 + 2/30 = 3/30 = 1/10, so f1 = 10 cm. With the concave lens, the image shifts by 30 cm, so the new image distance is 60 cm. 1/60 - 1/-15 = 1/F_eq => 1/60 + 4/60 = 5/60 = 1/12, so F_eq = 12 cm. Since 1/F_eq = 1/f1 + 1/f2, 1/12 = 1/10 + 1/f2 => 1/f2 = 1/12 - 1/10 = (5-6)/60 = -1/60. So f2 = -60 cm. The focal lengths are 10 cm and -60 cm.

Multiple choice luminous intensity measurements physics

A photographic plate is placed directly in front of a small diffused source in the sharp of a circular disc. It takes $12s$ to get a good exposure. If the source is rotated by $ { 60 }^{ \circ  }$ about one of its diameters, the time needed to get the same exposure will be

  1. $6 s$
  2. $12 s$
  3. $24 s$
  4. $48 s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let intensity of incident radiation $ = I _{0}$
This incident radiation passes for 12s 
This implies the exposure is $ 12I _{0}$
If the photographic plate is at an angle $\theta$
Radiation passing through is given by:
$I = I _{0} cos\theta$ Given that $\theta = 60^{\circ} C$
$I = \dfrac{I _{0}}{2}$
For the same exposure:
$12I _{0} = \dfrac{I _{0}}{2} \times t$
i.e. we get $t = 24s$
Option C is correct.

Multiple choice luminous intensity measurements physics

A photographic plate placed at a distance of $5 cm$ from a weak point source is exposed for $3 s$. if the plate is kept at a distance of $10 cm$ from the source, the time needed for the same exposure is

  1. $3 s$
  2. $12 s$
  3. $24 s$
  4. $48 s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The intensity is the power delivered per unit area, and hence it is inversely proportional to the square of the distance from source.


$I = \dfrac{k}{d^{2}}$

initially,

$I _{0} = \dfrac{k}{25}$ 

Exposure $ = 3I _{0} = \dfrac{3k}{25}$

Then distance is changed to 10cm

$I _{2} = \dfrac{k}{100}$

Time to get same exposure = t.

$\dfrac{kt}{100} = \dfrac{3k}{25}$

We get $t = 12s$
So, the answer is option (B).

Multiple choice evs reflection of light by plane surfaces characteristics of an image formed by a plane mirror formation of image in a plane mirror characteristics of the image formed by a plane mirror

An object is placed 15 cm from a diverging mirror, of radius of curvature 20 em. What is the image magnification produced?

  1. +0.4

  2. -0.4

  3. +2

  4. -2
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Answer is D.

For a spherically curved mirror in air, the magnitude of the focal length is equal to the radius of curvature of the mirror divided by two. The focal length is positive for a concave mirror.
Therefore, the focal length of the mirror is 20 cm / 2 = 10 cm.
The mirror equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:
1/f = 1/do + 1/di
So, 1/di = 1/f - 1/do = 1/10 - 1/15
Therefore, di = 30 cm.
The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:
M = hi / ho = -(di / do) = -(30 / 15) = -2.
Hence, the image magnification produced is -2.

Multiple choice evs reflection of light by plane surfaces characteristics of an image formed by a plane mirror formation of image in a plane mirror characteristics of the image formed by a plane mirror

The diameter of the sun subtends an angle of $ 0.5^0 $ at the surface of the earth. A converging lens of focal length 100 cm is used to provide an image of the sun on to a screen.the diameter ( in mm) of the image formed is nearly 

  1. 1

  2. 3

  3. 5

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice evs reflection of light by plane surfaces characteristics of an image formed by a plane mirror formation of image in a plane mirror characteristics of the image formed by a plane mirror

If $x _1$ be the size of the magnified image and $x _2$ the size of the diminished image in Lens Displacement Method, then the size of the object is

  1. $\sqrt{x _1x _2}$
  2. $x _1x _2$
  3. $x _1^2x _2$
  4. $x _1x^2 _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The size of the object in accordance with the Fresnel's displacement law is given by $\sqrt{x _1x _2}$.